RS Aggarwal Class 10 Solutions Chapter 10 - Quadratic Equations Ex 10D ( 10.4)

RS Aggarwal Class 10 Chapter 10 - Quadratic Equations Ex 10D ( 10.4) Solutions Free PDF

Students of Class 10 should have thorough knowledge about their topics and concepts of Class 10 maths so that they pass the exam with flying colors. So, students are advised to refer to the RS Aggarwal Solutions Class 10 Chapter 10 – Quadratic Equations Ex 10.4 which provide explanations for important topics and help students to solve their doubts in regard to difficult questions. These solutions are the best study material for Class 10 Maths and you can definitely rely on these solutions.

The RS Aggarwal solutions for Chapter 10 Ex 10.4 for Class 10 are prepared by subject experts in accordance to the latest syllabus of CBSE Class 10 Maths. The format of the RS Aggarwal Class 10 Solutions is easy to understand, which enable the students to learn each and every concept in quick time.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 10– Quadratic Equations 10D (10.4)

SOLVE EACH OF THE FOLLOWING EQUATIONS BY USING THE METHOD OF COMPLETING THE SQUARE:

QUESTION-1:

\(x^{2}-6x+3=0\)

Solution:

\(x^{2}-6x+3=0\)

\(\Rightarrow x^{2}-6x=-3\)

\(\Rightarrow x^{2}-2\times x\times 3+ 3^{2}=-3+3^{2}\) (Adding \(3^{2}\) on both sides)

\(\Rightarrow \left ( x-3 \right )^{2}=-3+9=6\)

\(\Rightarrow x-3 =\pm \sqrt{6}\) (Taking square root on both sides)

\(\Rightarrow x-3 = \sqrt{6}\:or\:x-3=-\sqrt{6}\)

\(\Rightarrow x=3+\sqrt{6}\:or\:x=3-\sqrt{6}\)

Hence, \(3+\sqrt{6}\:and\:3-\sqrt{6}\) are the roots of the given equation.

QUESTION-2:

\(x^{2}-4x+1=0\)

Solution:

\(x^{2}-4x+1=0\)

=> \(x^{2}-4x=-1\)

=> \(x^{2}-2\times x\times 2+2^{2}=-1+2^{2}\) (Adding \(2^{2}\) on both sides)

=> \(\left ( x-2 \right )^{2}=-1+4=3\)

=> \(\Rightarrow x-2 =\pm \sqrt{3}\) (Taking square root on both sides)

\(\Rightarrow x-2 = \sqrt{3}\:or\:x-2=-\sqrt{3}\)

\(\Rightarrow x=2+\sqrt{3}\:or\:x=2-\sqrt{3}\)

Hence, \(2+\sqrt{3}\:and\:2-\sqrt{3}\) are the roots of the given equation.

QUESTION-3:

\(x^{2}+8x-2=0\)

Solution:

=> \(x^{2}+8x=2\)

=> \(x^{2}+2\times x\times 4+4^{2}=2+4^{2}\) (Adding \(4^{2}\) on both sides)

=> \(\left ( x+4 \right )^{2}=2+16=18\)

=> \(x+4=\pm \sqrt{18}=\pm 3\sqrt{2}\) (Taking square root on both sides)

=> \(x+4= 3\sqrt{2}\:or\:x+4=-3\sqrt{2}\)

=> \(x=-4+ 3\sqrt{2}\:or\:x=-4-3\sqrt{2}\)

=> Hence, \(\left ( -4+3\sqrt{2} \right )\:and\:\left ( -4-3\sqrt{2} \right )\) are the roots of the given equation.

QUESTION-4:

\(4x^{2}+4\sqrt{3}x+3=0\)

Solution:

\(4x^{2}+4\sqrt{3}x+3=0\)

=> \(4x^{2}+4\sqrt{3}x=-3\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \sqrt{3}+\left ( \sqrt{3} \right )^{2}\) (Adding \(\left ( \sqrt{3} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\sqrt{3} \right )^{2}=-3+3=0\)

\(\Rightarrow 2x+\sqrt{3}=0\)

\(\Rightarrow x=-\frac{\sqrt{3}}{2}\)

Hence, \(-\frac{\sqrt{3}}{2}\) is the repeated root of the given equation.

QUESTION-5:

\(2x^{2}+5x-3=0\)

Solution:

\(2x^{2}+5x-3=0\)

\(\Rightarrow 4x^{2}+10x-6=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}+10x=6\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x+\frac{5}{2} \right )^{2}=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}=\left ( \frac{7}{2} \right )^{2}\)

\(\Rightarrow 2x+\frac{5}{2} =\pm \frac{7}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x+\frac{5}{2} = \frac{7}{2}\:or\:2x+\frac{5}{2}=-\frac{7}{2}\)

\(\Rightarrow 2x=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1\:or\:2x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6\)

=> x = \(\frac{1}{2}\) or x = -3

Hence, \(\frac{1}{2}\) and -3 are the roots of the given equation.

QUESTION-6:

\(3x^{2}-x-2=0\)

Solution:

\(3x^{2}-x-2=0\)

\(9x^{2}-3x-6=0\)

\(9x^{2}-3x=6\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 3x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=6+\left ( \frac{1}{2} \right )^{2}\) [Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 3x-\frac{1}{2} \right )^{2}=6+\frac{1}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow 3x-\frac{1}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow 3x-\frac{1}{2}=\frac{5}{2}\:or\:3x-\frac{1}{2}=-\frac{5}{2}\)

\(\Rightarrow 3x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\:or\:3x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2\)

=> x = 1 or x = \(-\frac{2}{3}\)

Hence, 1 and \(-\frac{2}{3}\) are the roots of the given equation.

QUESTION-7:

\(8x^{2}-14x-15=0\)

Solution:

\(8x^{2}-14x-15=0\)

\(\Rightarrow 16x^{2}-28x-30=0\) (Multiplying both sides by 2)

\(\Rightarrow 16x^{2}-28x=30\)

\(\Rightarrow \left ( 4x \right )^{2}-2\times 4x\times \frac{7}{2}+\left ( \frac{7}{2} \right )^{2}=30+\left ( \frac{7}{2} \right )^{2}\) [Adding \(\left ( \frac{7}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 4x-\frac{7}{2} \right )^{2}=30+\frac{49}{4}=\frac{169}{4}=\left ( \frac{13}{2} \right )^{2}\)

\(\Rightarrow 4x-\frac{7}{2}=\pm \frac{13}{2}\) (Taking square root on both sides)

\(\Rightarrow 4x-\frac{7}{2}=\frac{13}{2}\:or\:4x-\frac{7}{2}=-\frac{13}{2}\)

\(\Rightarrow 4x=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10\:or\:4x=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3\)

=> x = \(\frac{5}{2}\) or x = \(-\frac{3}{4}\)

Hence, \(\frac{5}{2}\) and \(-\frac{3}{4}\) are the roots of the given equation.

QUESTION-8:

\(7x^{2}+3x-4=0\)

Solution:

\(7x^{2}+3x-4=0\)

\(\Rightarrow 49x^{2}+21x-28=0\) (Multiplying both sides by 7)

\(\Rightarrow 49x^{2}+21x=28\)

\(\Rightarrow \left ( 7x \right )^{2}+2\times 7x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=28+\left ( \frac{3}{2} \right )^{2}\) [Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 7x+\frac{3}{2} \right )^{2}=28+\frac{9}{4}=\frac{121}{4}=\left ( \frac{11}{2} \right )^{2}\)

\(\Rightarrow 7x+\frac{3}{2}=\pm \frac{11}{2}\) (Taking square root on both sides)

\(\Rightarrow 7x+\frac{3}{2}=\frac{11}{2}\:or\:7x+\frac{3}{2}=-\frac{11}{2}\)

\(\Rightarrow 7x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4\:or\:7x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7\)

=> x = \(\frac{4}{7}\) or x = -1

Hence, \(\frac{4}{7}\) and -1 are the roots of the given equation.

QUESTION-9:

\(3x^{2}-2x-1=0\)

Solution:

\(3x^{2}-2x-1=0\)

\(\Rightarrow 9x^{2}-6x-3=0\) (Multiplying both sides by 3)

\(\Rightarrow 9x^{2}-6x=3\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 1+1^{2}=3+1^{2}\) [Adding \(1^{2}\) on both sides]

\(\Rightarrow \left (3x-1 \right)^{2}=3+1=4=\left ( 2 \right )^{2}\)

\(\Rightarrow 3x-1 =\pm 2\) (Taking square root on both sides)

=> 3x – 1 = 2 or 3x – 1 = -2

=> 3x = 3 or 3x = -1

=> x = 1 or x = \(-\frac{1}{3}\)

Hence, 1 and \(-\frac{1}{3}\) are the roots of the given equation.

QUESTION-10:

\(5x^{2}-6x-2=0\)

Solution:

\(5x^{2}-6x-2=0\)

\(\Rightarrow 25x^{2}-30x-10=0\) (Multiplying both sides by 5)

\(\Rightarrow 25x^{2}-30x=10\)

\(\Rightarrow \left ( 5x \right )^{2}-2\times 5x\times3+3^{2}=10+3^{2}\) [Adding \(3^{2}\) on both sides]

\(\Rightarrow \left ( 5x-3 \right )^{2}=10+9=19\)

\(\Rightarrow 5x-3=\pm \sqrt{19}\) (Taking square root on both sides)

\(\Rightarrow 5x-3= \sqrt{19}\:or\:5x-3=-\sqrt{19}\)

\(\Rightarrow 5x=3+\sqrt{19}\:or\:5x=3-\sqrt{19}\)

\(\Rightarrow x=\frac{3+\sqrt{19}}{5}\:or\:x=\frac{3-\sqrt{19}}{5}\)

Hence, \(\frac{3+\sqrt{19}}{5}\:and\:\frac{3-\sqrt{19}}{5}\) are the roots of the given equation.

QUESTION-11:

\(\frac{2}{x^{2}}-\frac{5}{x}+2=0\)

Solution:

\(\Rightarrow \frac{2-5x+2x^{2}}{x^{2}}=0\)

\(\Rightarrow 2x^{2}-5x+2=0\)

\(\Rightarrow 4x^{2}-10x+4=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}-10x=-4\)

\(\Rightarrow \left ( 2x \right )^{2}-2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}=-4+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x-\frac{5}{2} \right )^{2}=-4+\frac{25}{4}=\frac{9}{4}=\left ( \frac{3}{2} \right )^{2}\)

\(\Rightarrow 2x-\frac{5}{2} = \pm \frac{3}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x-\frac{5}{2}=\frac{3}{2}\:or\:2x-\frac{5}{2}=-\frac{3}{2}\)

\(\Rightarrow 2x = \frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4\:or\:2x=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1\)

=> x = 2 or x = \(\frac{1}{2}\)

Hence, 2 and \(\frac{1}{2}\) are the roots of the given equation.

QUESTION-12:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

Solution:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

\(\Rightarrow 4x^{2}+4bx=a^{2}-b^{2}\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times b+b^{2}=a^{2}-b^{2}+b^{2}\) [Adding \(b^{2}\) on both sides]

\(\Rightarrow \left ( 2x+b \right )^{2}=a^{2}\)

\(\Rightarrow 2x+b =\pm a\) (Taking square root on both sides)

=> 2x+b = a or 2x+b = -a

=> 2x = a-b or 2x = -a-b

\(\Rightarrow x=\frac{a-b}{2}\:or\:x= -\frac{a+b}{2}\)

Hence, \(\frac{a-b}{2}\:and\:-\frac{a+b}{2}\) are the roots of the given equation.

Question-13:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

Solution:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=-\sqrt{2} \)

\(\Rightarrow x^{2}-2\times x\times \left ( \frac{\sqrt{2}+1}{2} \right )+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}=-\sqrt{2}+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) [Adding \(\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left [ x-\left ( \frac{\sqrt{2}+1}{2}^{2} \right ) \right ]^{2}=\frac{-4\sqrt{2}+2+1+2\sqrt{2}}{4}=\frac{2-2\sqrt{2}+1}{4}=\left ( \frac{\sqrt{2}-1}{4} \right )^{2}\)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\pm \left ( \frac{\sqrt{2}-1}{2} \right )\) (Taking square root on both sides)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\left ( \frac{\sqrt{2}-1}{2} \right )\:or\:x-\left ( \frac{\sqrt{2}+1}{2} \right )=-\left ( \frac{\sqrt{2}-1}{2} \right )\)

\(\Rightarrow x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\:or\:x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}\)

\(\Rightarrow x=\frac{2\sqrt{2}}{2}=\sqrt{2}\:or\:x=\frac{2}{2}=1\)

Hence, \(\sqrt{2}\) and 1 are the roots of the given equation.

Question-14:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

Solution:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x-4=0\) (Multiplying both sides by \(\sqrt{2}\))

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x=4\)

\(\Rightarrow \left ( \sqrt{2}x \right )^{2}-2\times \sqrt{2}x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=4+\left ( \frac{3}{2} \right )^{2}\) (Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( \sqrt{2}x-\frac{3}{2} \right )^{2}=4+\frac{9}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\frac{5}{2}\:or\:\sqrt{2}x-\frac{3}{2}=-\frac{5}{2}\)

\(\Rightarrow \sqrt{2}x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\:or\:\sqrt{2}x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1\)

\(\Rightarrow x=\frac{4}{\sqrt{2}}=2\sqrt{2}\:or\:x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\)

Hence, \(2\sqrt{2}\:and\:-\frac{\sqrt{2}}{2}\) are the roots of the given equation.

Question-15:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

Solution:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x+21=0\) (Multiplying both sides by \(\sqrt{3}\))

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x=-21\)

\(\Rightarrow \left ( \sqrt{3x} \right )^{2}+2\times \sqrt{3}x\times 5+5^{2}=-21+5^{2}\) (Adding \(5^{2}\) on both sides)

=> \(\left ( \sqrt{3}x+5 \right )^{2}=-21+25=4=2^{2}\)

=> \(\sqrt{3}x+5=\pm 2\) (Taking square root on both sides)

=> \(\sqrt{3}x+5=2\:or\:\sqrt{3}x+5=-2\)

=> \(\sqrt{3}x=-3\:or\:\sqrt{3}x=-7\)

=> x = \(-\frac{3}{\sqrt{3}}=-\sqrt{3}\:or\:x=-\frac{7}{\sqrt{3}}=-\frac{7\sqrt{3}}{3}\)

Hence, \(-3\:and\:-\frac{7\sqrt{3}}{3}\) are the roots of the given equation.

Question-16:

By using the method of completing the square, show that the equation \(2x^{2}+x+4=0\) has no real roots.

Solution:

\(2x^{2}+x+4=0\)

=> \(4x^{2}+2x+8=0\) (Multiplying both sides by 2)

=> \(4x^{2}+2x=-8\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=-8+\left ( \frac{1}{2} \right )^{2}\) (Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\frac{1}{2} \right )^{2}=-8+\frac{1}{4}=-\frac{31}{4}< 0\)

But, \(\left ( 2x+\frac{1}{2} \right )^{2}\) cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.

Key Features of RS Aggarwal Class 10 Solutions Chapter 10 – Quadratic Equations Ex 10D ( 10.4)

  • The solutions provided here help students to perform well in their exams.
  • It provides students with variety of questions to practice.
  • By solving the RS Aggarwal Maths solutions you can clear your doubts.
  • It is the best material for revising the entire syllabus.

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How many electrons a carbon atom shares with other carbon atom in the formation of acetylene molecule?

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